# prove that $f^{-1} = f*$ as follow

edit:

Let $$f: A \to B$$. Let $$f*$$ be the inverse relation, i.e. $$\begin{equation*} f* = \{(y,x) \in B \times A \mid f(x)=y \}. \end{equation*}$$ Show that if $$f* : B \to A$$ is a function, then $$f^{-1} = f*$$.

Attempt:

Let $$f*: B \to A$$ be a function. Then, $$f$$ is bijective and hence $$f$$ is invertible, i.e. $$f$$ have an inverse, say $$f^{-1}$$. It is clear that $$f^{-1} \circ f = i_A$$ and $$f \circ f* = i_B$$ where $$i$$ is the identity function. Hence, $$\begin{equation*} f^{-1} = f^{-1} \circ i_B = f^{-1} \circ (f \circ f*) = (f^{-1} \circ f) \circ f* = i_A \circ f* = f*. \end{equation*}$$ Thus, $$f^{-1} = f$$.

Is the above correct?

• Since $i = f^{-1}\circ f : A \to A$ is the identity function on $A$, how is defined $f^{-1}\circ i$? Also, $f \circ f^*$ is the identity function on $B$, it is not the same as $i$. Commented Jan 6, 2021 at 7:28
• I assume your defintion of $f^{-1}$ is "any left-inverse"? And instead of "It is clear that", you may argue more explicitly where you use the fact that $f^*$ is a function. Commented Jan 6, 2021 at 7:29
• Suggestion for improvement: Compute the relation(!) compositions $f^*\circ f\subseteq A\times A$ and $f\circ f^*\subseteq B\times B$ and see what happens when both $f$ and $f^*$ are functions Commented Jan 6, 2021 at 7:32
• The question is missing a proper definition of $f^{-1}$. I will not settle for "the usual definition of the inverse".
– user65203
Commented Jan 6, 2021 at 7:52
• What is the connection of the title with the question ?
– user65203
Commented Jan 6, 2021 at 7:54

First, suppose that $$f*$$ is a function. That means that each for each $$y \in B, \exists!x \in A$$ such that $$f^*(y)=x.$$ Then $$(f^* \circ f)(x) = f^*(f(x)) = f^*(y)=x,$$ and similarly, $$(f \circ f^*)(y) =f(f^*(y))= f(x)=y$$, so $$f*$$ is the inverse of $$f$$. Or perhaps I should say $$f*$$ is an inverse of $$f$$, since you've also been tasked with proving the uniqueness of the inverse.
To prove that the inverse of f is unique, suppose that f has two inverses, g and h. We need to show that g=h. Since g is the inverse of f, that means $$g∘f=id_A$$ and $$f∘h=id_B$$. Then $$g=g∘id_B=g∘(f∘h)=(g∘f)∘h=id_A∘h=h$$.
• but, how to prove $f^{-1} = f*$. Commented Jan 6, 2021 at 10:35
• Yes, the basic proof that $f*$ is an inverse of $f$ is correct, you just needed the uniqueness part. And at the end, did you mean to put "Thus, $f^{-1} = f^*$"? Commented Jan 6, 2021 at 10:47